// QUANT INTERVIEW PREP
Expected Value Quant Interview Prep & Practice
Expected value is how trading firms ask whether you would take a bet. The questions below price games, wagers and payoffs the way a market-making interview does.
Sample Expected Value Questions
Question 1 · Optiver Trading Assessment
You win 10 with probability 55% and lose 10 otherwise. Expected value per bet?
- 0
- 5.5
- 1
- 10
Question 2 · SIG Problem-Solving Assessment
In a room of 30 people with birthdays independent and uniform over 365 days, what is the expected number of PAIRS of people who share a birthday?
See the solution ↓Question 3 · Akuna Grid Game
A 10-sided die is rolled until it shows a 1. The first 1 on roll n pays $10^n. The game is stopped after 4 rolls and pays nothing if no 1 has come up. What is the expected payout?
See the solution ↓Question 4 · Five Rings Quant Trading Screen
7,900 orders are routed independently and uniformly at random across 7,900 venues. How many venues expect to receive nothing?
See the solution ↓Question 5 · Old Mission Quant Screen
16 traders leave 16 identical-looking coats at the door and each takes one at random on the way out. What is the expected number who get their own coat?
- 16
- 1
- 0.06
- 8
Step-by-Step Solutions
Solution 1
You win 10 with probability 55% and lose 10 otherwise. Expected value per bet?
Answer: 1
55% × 10 − 45% × 10 = 1. The edge is the probability gap times the stake.
Solution 2
In a room of 30 people with birthdays independent and uniform over 365 days, what is the expected number of PAIRS of people who share a birthday?
Answer: 1.192
Each of the C(30,2) = 435 pairs matches with probability 1/365, so by linearity the expectation is 435/365 = 1.19. No independence between pairs is needed - linearity does not care.
Solution 3
A 10-sided die is rolled until it shows a 1. The first 1 on roll n pays $10^n. The game is stopped after 4 rolls and pays nothing if no 1 has come up. What is the expected payout?
Answer: 820
The first 1 lands on roll n with probability (9/10)^(n-1) x 1/10, and that roll pays $1 x 10^n. Multiplying gives $1 x 9^(n-1), so the terms grow rather than shrink and the sum over the 4 rolls is $820. Same shape as St Petersburg: the payout outruns the probability, so only the cap keeps the expectation finite.
Solution 4
7,900 orders are routed independently and uniformly at random across 7,900 venues. How many venues expect to receive nothing?
Answer: 2906
A given venue is missed with probability (1 − 1/7,900)^7,900, which is 1/e ≈ 0.368. By linearity, 7,900 × 0.368 ≈ 2907.
Solution 5
16 traders leave 16 identical-looking coats at the door and each takes one at random on the way out. What is the expected number who get their own coat?
Answer: 1
Each trader gets their own coat with probability 1/16, and expectation adds regardless of the dependence between them: 16 × 1/16 = 1. The answer does not move with 16, which is the point of the question.
How to Pass the Expected Value Screen
Lay out every outcome with its probability and its payoff, then sum. Linearity of expectation lets you split a complicated payoff into simple pieces.
Separate the fair price from the decision. The EV tells you what the game is worth; whether you play depends on the price you are offered against it.
Watch for the stake. Net profit and gross return differ by the cost of playing, and screens test exactly that difference.
Practice expected value alongside 4,000+ more procedurally generated scenarios under live timer conditions. Unlock Outcry Pro.
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